A motorcycle and a car start from rest from the same place at the same time and travel in the same direction. The motorcycle accelerates at 1.0 m $s^{- 2}$ up to a speed of 36 km $h^{- 1}$ and the car at
0.5 m $s^{- 2}$ up to a speed of 54 km $h^{- 1}$ . The time at which the car would overtake the motorcycle is

20 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

35 s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $35 s$
As the car overtakes the motorcycle, the distance covered by them will be same. Let the total distance travelled be $S$ and time taken while overtaking is $t$ .

Now for the motor cycle the maximum speed attained is given as:
$36 k m / h = 36 \times \frac{5}{18} = 10 m / s$

Given that the acceleration of bike is $= 1.0 m / s^{2}$
the motor cycle takes time $t_{1}$ to attain the maximum speed and it is given by:
$v = u + a t_{1} \Rightarrow 10 = 0 + 1.0 \times t_{1}$

$t_{1} = 10 s$ (since u = 0)

The distance covered by motocycle in attaining the maximum speed is
$S_{1} = 0 + \frac{1}{2} a t_{1}^{2} = \frac{1}{2} \times 1.0 \times (10)^{2} = 50 m$

The time during which the motorcycle moves with maximum speed is $(t - 10) s$ .
The distance covered by the motorcycle during the time is
$S'_{1} = 10 \times (t - 10)$
$= (10 t - 100) S'_{1} = 10 \times (t - 10) = (10 t - 100) m$

$∴$ Total distance traveled by motorcycle in time t is
$S = S_{1} + S'_{1}$
$= 50 + (10 t - 100)$
$= (10 t - 50) m$ ....... (i)

Now similarly for car :
Maximum speed attained by the car is
$= 54 k m / h = 54 \times \frac{5}{18} = 15 m / s$

Since its acceleration of the car is $= 0.5 m / s^{2}$
The time taken by it to attain the maximum speed is given by
$15 = 0 + 0.5 \times t_{2}$ or $t_{2} = 30 s$ (since u = 0)

The distance covered by the car in attaining the maximum speed is
$S_{2} = 0 + \frac{1}{2} a t_{2}^{2} = 12 \times 0.5 \times (30)^{2}$
$= 225 m$

The time during which the car moves with maximum speed is (t - 30) s.
The distance covered by the car during this time is
$S'_{2} = 15 \times (t - 30) = (15 t - 450)$

$∴$ Total distance travelled by car in time $t$ is
$S = S_{2} + S'_{2} = 225 + (15 t - 450) = (15 t - 225) m$ ...(ii)

From equations (i) and (ii), we get
$10 t - 50 = 15 t - 225$
$10 t - 50 = 15 t - 225$
$\Rightarrow 5 t = 175$
$\Rightarrow t = 35 s$

Suggest Corrections

Similar questions

Q. A motorcycle travelling on the highway at a speed of

120 k m / h

passes a car travelling at a speed of 90 km/h. From the point of view of a passenger on the car, what is the velocity of the motorcycle?

Q. A

200

metre long train passes a motorcycle, running in the same direction at

12

kmph, in

15

seconds and a car travelling in the same direction in

20

seconds. At what speed is the car travelling (length of both the motorcycle and car is negligible)?

Robbers in a car travelling at 20m/s pass a policeman on a motorcycle at rest. The policeman immediately starts chasing the robbers. The policeman accelerates at 3m/s² for 12s and thereafter travels at constant velocity. Calculate the distance covered by the policeman before he overtakes the car.

Q. A motorcyclist situated at the origin is located at a distance of 12 m behind a car. At t = 0, the motorcyclist starts moving with a constant velocity v = 8 m/s and same time the car starts accelerating from rest with a =

m s^{- 2}

. (a) When and where do they meet? (b) Draw the position-time and velocity-time relations of car and motorcycle on the same graph.

A car traveling at 20 km/h speeds up to 60 km/h in 6 seconds. What is its acceleration in m/ $s^{2}$ ?

Join BYJU'S Learning Program

A motorcycle and a car start from rest from the same place at the same time and travel in the same direction. The motorcycle accelerates at 1.0 m s−2 up to a speed of 36 km h−1 and the car at 0.5 m s−2 up to a speed of 54 km h−1. The time at which the car would overtake the motorcycle is